Question

Special Relativity

A photon of energy E collides with a stationary particle of rest mass m0 and is absorbed. What is the velocity of the resulting composite particle?

NOTE: I used consrvation of momentum: p(before) = p(after), therefore p/c (energy of photon) = gamma*m0*v and solved for v. Apparently this is incorrect????

Answer 1

momentum of photon is P = (E/C) momentum consrvation says that Total momentum before collission is equal to total momentum after collision (E/C) = r M v v/(Squareroot(1 - (v^2/c^2))) = E/M_0 c: (1- (v^2/c^2)/v^2)c = (M_0^2 C^2/E^2) c^2 - v^2/c^2v^2 = (M_0^2 c^4/E^2) c^2/v^2 - 1 = (M_0^2 c^4/E^2) \r\n c^2/v^2 = 1 + (M_0^2 c^4/E^2) c^2/v^2 = (E^2 + M_0^2 C^4/E^2) V^2/c^2 = (E^2/E^2 + M_0^2 c^4) V^2 = (c^2 E^2/E^2 + M_0^2 C^4) V = Squareroot(c^2E^2/E^2 + M_0^2 C^4) is the velocity of the composite particle